How conditions affect K, and ΔStotal = R ln K

What changes K (spec 13.4 to 13.7)

• Concentration or pressure: K is unchanged. The system shifts until the concentrations again fit the same K. For example, raising the pressure on 2SO2 + O2 ⇌ 2SO3 makes the quotient temporarily smaller than Kp, so more SO3 forms until it equals Kp again.
• Catalyst: K is unchanged; equilibrium is reached faster.
• Temperature: K changes. For an exothermic forward reaction, K decreases as temperature rises; for an endothermic forward reaction, K increases. The shift in position is explained by the change in K.

Entropy and K (spec 13.8)

ΔStotal = R ln K

• K > 1 gives ΔStotal > 0 (products favoured).
• K < 1 gives ΔStotal < 0.

Raising the temperature makes ΔSsurroundings = -ΔH/T smaller in size. For an exothermic reaction (positive ΔSsurroundings) ΔStotal falls, so K falls; for an endothermic reaction ΔStotal rises, so K rises.

Extent of reaction (spec 13.9)

• K greater than about 10¹⁰: reaction essentially goes to completion.
• K less than about 10^-10: essentially no reaction.
• K between these: a mixture with significant amounts of reactants and products.

Worked example

For a reaction, K = 1.0 x 10¹⁰ at 298 K. Calculate ΔStotal.

1. ΔStotal = R ln K = 8.31 x ln(1.0 x 10¹⁰)
2. ln(1.0 x 10¹⁰) = 23.03

Answer: ΔStotal = +191 J mol^-1 K^-1, so the reaction is essentially complete

Check yourself

1. For an exothermic reaction, what happens to K when the temperature is raised?
   • K decreases
   • K increases
   • K stays the same
   • K becomes zero
   Show answer

   K decreases. Higher temperature favours the endothermic backward reaction, which means a smaller K.

2. The total pressure on an equilibrium mixture of gases is increased at constant temperature. What happens to Kp?
   • It increases
   • It decreases
   • It stays the same
   • It doubles
   Show answer

   It stays the same. Pressure changes move the position, not the value of Kp.

3. How does adding a catalyst affect an equilibrium?
   • Equilibrium is reached faster but K and the position are unchanged
   • K increases
   • The position shifts to the right
   • K decreases
   Show answer

   Equilibrium is reached faster but K and the position are unchanged. A catalyst speeds up the forward and backward reactions equally.

4. Calculate ΔStotal for a reaction with K = 1.0 x 10¹⁰. (R = 8.31 J mol^-1 K^-1)
   Show answer

   +191 J mol^-1 K^-1. ΔStotal = 8.31 x ln(1.0 x 10¹⁰) = 8.31 x 23.03 = +191 J mol^-1 K^-1.

5. A reaction has ΔStotal = -20.0 J mol^-1 K^-1. Calculate K. (R = 8.31 J mol^-1 K^-1)
   Show answer

   0.0901. ln K = -20.0 / 8.31 = -2.407; K = e^-2.407 = 0.0901.

6. A reaction has K = 3 x 10¹⁵ at room temperature. What can you conclude?
   • Almost no products form
   • The reaction is fast
   • The reaction is endothermic
   • The reaction goes essentially to completion
   Show answer

   The reaction goes essentially to completion. A very large K means products dominate, but K says nothing about rate.

7. The forward reaction of an equilibrium is endothermic. Why does K increase with temperature?
   • The activation energy decreases
   • ΔSsurroundings becomes less negative, so ΔStotal and therefore K increase
   • More reactant molecules form
   • ΔSsystem becomes zero
   Show answer

   ΔSsurroundings becomes less negative, so ΔStotal and therefore K increase. ΔSsurr = -ΔH/T is negative and shrinks in size as T rises, so ΔStotal = R ln K rises.

Practise with instant feedback Print this lesson

Sources

• Pearson Edexcel IAL Chemistry (2018) specification
• Pearson Edexcel IAL Chemistry (2018) scheme of work

Last updated 7 November 2026. Spotted a mistake? Tell us.